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Chapter 20: Problem 65

A volume of \(6.00 \mathrm{~L}\) of a monatomic ideal gas, originally at \(400 . \mathrm{K}\) and a pressure of \(3.00 \mathrm{~atm}\) (called state 1 ), undergoes the following processes, all done reversibly: \(1 \rightarrow 2\) isothermal expansion to \(V_{2}=4 V_{1}\) \(2 \rightarrow 3\) isobaric compression \(3 \rightarrow 1\) adiabatic compression to its original state Find the entropy change for each process.

Short Answer

Expert verified
Question: Calculate the entropy change for each process that a monatomic ideal gas undergoes during the following three process cycle: 1) Isothermal expansion from V1 to V2 = 4 * V1, 2) Isobaric compression, and 3) Adiabatic compression back to the original state. The initial conditions of the gas are P1 = 3.00 atm, V1 = 6.00 L, and T1 = 400 K. Answer: The entropy changes for each process are given by: 1) Isothermal expansion: ΔS₁→₂ = nR ln(4) 2) Isobaric compression: ΔS₂→₃ = n(5/2)R ln(T₂/T₁) 3) Adiabatic compression: ΔS₃→₁ = 0

Step by step solution

01

Isothermal expansion

To find the entropy change in an isothermal process, we can use the formula: \(\Delta S = nR\ln \left(\frac{V_2}{V_1}\right)\) We first need to find the number of moles, n, using the initial conditions of the gas: \(P_1 = 3.00\ \mathrm{atm},\ V_1 = 6.00\ \mathrm{L},\ T_1 = 400\ \mathrm{K}\) Using the ideal gas law \(PV=nRT\), we have: \(n = \frac{P_1V_1}{RT_1} = \frac{3.00\ \mathrm{atm} \cdot 6.00\ \mathrm{L}}{0.0821\ \mathrm{L\ atm/mol\ K} \cdot 400\ \mathrm{K}}\) Now, we can find the entropy change in the isothermal expansion: \(\Delta S_{1\rightarrow2} = nR\ln \left(\frac{V_2}{V_1}\right) = nR\ln(4)\)
02

Isobaric compression

For an isobaric process, the formula for the entropy change is: \(\Delta S = nC_P\ln \left(\frac{T_2}{T_1}\right)\) Since the process is isobaric, the pressure at state 2 is the same as state 1. We can use the ideal gas law to determine the temperature at state 2: \(P_1V_1 = nRT_1 = nRT_2\) Rearranging for T2 gives: \(T_2 = \frac{P_1V_1}{nR} = \frac{3.00\ \mathrm{atm} \cdot 6.00\ \mathrm{L}}{0.0821\ \mathrm{L\ atm/mol\ K}}\) Because the gas is monatomic, we have: \(C_P = \frac{5}{2}R\) Now, we can find the entropy change in the isobaric compression: \(\Delta S_{2\rightarrow3} = nC_P\ln \left(\frac{T_2}{T_1}\right) = n\left(\frac{5}{2}R\right)\ln\left(\frac{T_2}{T_1}\right)\)
03

Adiabatic compression

In an adiabatic process, the entropy change is \(0\) since there is no heat transfer: \(\Delta S_{3\rightarrow1} = 0\) Finally, we calculate the entropy change for each process using the equations derived in the previous steps: \(\Delta S_{1\rightarrow2} = nR\ln(4)\) \(\Delta S_{2\rightarrow3} = n\left(\frac{5}{2}R\right)\ln\left(\frac{T_2}{T_1}\right)\) \(\Delta S_{3\rightarrow1} = 0\) Using these entropy changes, we are able to describe the entropy changes in the three processes which the monatomic ideal gas undergoes.

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Most popular questions from this chapter

The temperature at the cloud tops of Saturn is approximately 50. K. The atmosphere of Saturn produces tremendous winds; wind speeds of \(600 . \mathrm{km} / \mathrm{h}\) have been inferred from spacecraft measurements. Can the wind chill factor on Saturn produce a temperature at (or below) absolute zero? How, or why not?

The change in entropy of a system can be calculated because a) it depends only on the c) entropy always increases. initial and final states. d) none of the above. b) any process is reversible.

If the Earth is treated as a spherical black body of radius \(6371 \mathrm{~km}\), absorbing heat from the Sun at a rate given by the solar constant (1370. W/m \(^{2}\) ) and immersed in space with an approximate temperature of \(T_{s p}=50.0 \mathrm{~K},\) it radiates heat back into space at an equilibrium temperature of \(278.9 \mathrm{~K}\). (This is a slight refinement of the model in Chapter 18.) Estimate the rate at which the Earth gains entropy in this model.

A refrigerator with a coefficient of performance of 3.80 is used to \(\operatorname{cool} 2.00 \mathrm{~L}\) of mineral water from room temperature \(\left(25.0^{\circ} \mathrm{C}\right)\) to \(4.00^{\circ} \mathrm{C} .\) If the refrigerator uses \(480 . \mathrm{W}\) how long will it take the water to reach \(4.00^{\circ} \mathrm{C}\) ? Recall that the heat capacity of water is \(4.19 \mathrm{~kJ} /(\mathrm{kg} \mathrm{K}),\) and the density of water is \(1.00 \mathrm{~g} / \mathrm{cm}^{3}\). Assume the other contents of the refrig. erator are already at \(4.00^{\circ} \mathrm{C}\).

A heat engine consists of a heat source that causes a monatomic gas to expand, pushing against a piston, thereby doing work. The gas begins at a pressure of \(300 . \mathrm{kPa}\), a volume of \(150 . \mathrm{cm}^{3}\), and room temperature, \(20.0^{\circ} \mathrm{C}\). On reaching a volume of \(450 . \mathrm{cm}^{3}\), the piston is locked in place, and the heat source is removed. At this point, the gas cools back to room temperature. Finally, the piston is unlocked and used to isothermally compress the gas back to its initial state. a) Sketch the cycle on a \(p V\) -diagram. b) Determine the work done on the gas and the heat flow out of the gas in each part of the cycle. c) Using the results of part (b), determine the efficiency of the engine.
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